groups-rings-fields

[rocky cloak]

So for example
(3 4 5) (1 2 3) (3 4 5)^-1 = (1 2 4)

(The 3 was replaced by a 4)

[vestal sapphire]

Yeah we got a hint to this problem to find s^-1 for one of the cycles, but I couldnt get rid off those sts^-1 on the sides

[rocky cloak]

Get rid of them?

[vestal sapphire]

well so for the previous problem I was able to show that we can write pi = s_1 t_1 s_1^-1 ..., where tall t_i were in the form (1 2 j), but the s's werent' so I had to get rid off them somehow, but couldnt

its in polish

but basically applying this idea

[thorn jay]

S_3 of order three?

3 = 0

Omg

[amber burrow]

Show that if G is a nonabelian finite group, then $|Z(G)| \leqslant
\frac{|G|}{4}$

[cloud walrus]

donut123

[amber burrow]

i know how to solve, but i thought i was kinda cool, so wanted to share

[thorn jay]

||if |Z(G)| is either |G| or |G|/2 or |G|/3 (these are the only possibilities by Langarange) then |G/Z(G)| <= 3 so must be cyclic, hence G is abelian. In particular, this means that G/Z(G) must 'at least' be the Klein-four group||

[amber burrow]

yup, same solution!

[thorn jay]

Based

Never thought I'd use that fact ever

[candid patrol]

order doesn't necessary says that |G| = order

[tough raven]

In fairness, you don't know if you'll use this one either.

Or do you?

Hmmmmm.

[candid patrol]

Like to me, the dihedral group of order n is D2n

[tough raven]

I wonder if this statement can be seen as a special case of group extension theory.

[tough raven]

I would be more comfortable with saying that the dihedral group of order 2n is D_n lol.

[thorn jay]

I use order for that though, thats the problem, i guess

[tough raven]

So the theory says that such extensions are classified by H^2(C; Z) where C is the cyclic quotient and Z the central subgroup, viewed as a trivial C-module. This is not trivial (there are non-trivial extensions) but it might still be possible to show that every extension is abelian.

[thorn jay]

I use "dihedral group of order 2n is D_2n"

[thorn jay]

Use case enough
Im sure its useful somewhere

[tough raven]

That is the standard meaning of order. You have to say somewhere that you're using order with a meaning where order(S_n) = n or you can't expect that sentence to be understood.

[tough raven]

kek

[thorn jay]

I suppose in this case it can be inferred both from notation and from context

But yeah i agree

[amber burrow]

I wouldn’t say it’s super uncommon

But I was like trying to use class equation

There was another cool one

If M is a maximal subgroup of G, and M is normal, then |G:M| is prime

[thorn jay]

|G:M| = |G/M| right? In this case at least

||Then it follows from the fact that G/M must not have any subgroups, hence is prime cyclic||

[vapid vale]

major brainfart.. if S \subset R is the annihilator of N \subset M an R-module, is there a word for what N is in terms of S

[amber burrow]

Uh basically but you need some things

Let p some prine dividing |P:M|

By Cauchy theorem, there is a subgroup H of order p

By the fourth isomorphism theorem, this subgroup is of the form G/N for some N containing M

[south patrol]

Wait what am I thinking of lol

[amber burrow]

Was about to say 3/8 bigger

Than 1/4

[south patrol]

Ah sorry I'm thinking of this: the probability that two random elements in a non-abelian group commute is <= 5/8 I think

[amber burrow]

But M is maximal, so either N = M or N= G

[amber burrow]

That’s kinda interesting

[south patrol]

I also wonder like is this bound sharp?

I imagine it is not but am unsure lol

Oh, I guess D_8 is a counterexample lol

[amber burrow]

Klein 4

Yeah

[south patrol]

Klein 4 is abelian lol

But yes for the quotient nvm sure

Indeed non-trivial finite p groups have nontrivial centre

5/8 is similarly sharp for my question

I think you can just check with small groups like Q8

Every time I say D8 I brace myself for someone correcting me with D4

[amber burrow]

I was about to lmao

Why would anyone use D_2n

If you do, you should be obligated to write S_n!

[glad osprey]

and ∞Z for all the subgroups of Z

[tropic obsidian]

what does it mean for a pairing to be perfect?

is it just when a bilinear map is nondegenerate

[wraith cargo]

So you have a pairing A(x)B -> C
It's perfect when the associated map A-> C(x)B* and B->C(x)A* are isomorphisms

[tropic obsidian]

so it is the same as non degenerate?

[rocky cloak]

Non-degenerate means the maps are injective.

[tropic obsidian]

oh, so only in fin dim then

[rocky cloak]

Yeah, for findim vector spaces injective would imply isomorphisms (in the case C = k)

[thorn jay]

I mean yeah but thats pretty obvious right?
Let f : G -> G/M be your projection and suppose H < G/M a nontrivial proper subgroup, then f^{-1}(H) contains M, but cant be G as f is surjective. Thus, contradiction and G/M must contain no subgroups.
Then take any nonidentity element gM in G/M. As it has no subgroups <gM> = G/M, so it is simple and cyclic, so prime cyclic

[thorn jay]

I do because i learned it and didnt stop

:3c

[hidden wind]

help

that’s so clunky

[thorn jay]

S_120, the group with 120 elements

Obviously

[hidden wind]

pfft

is that… the full icosahedral group? the binary icosahedral group? isn’t there another one even?

oh S5

😭

[thorn jay]

A_60 the famous smallest nonabelian simple group

[hidden wind]

the icosahedral group

[neat zodiac]

is there a trick to this question or must it be done manually

[barren sierra]

there's a good formula for conjugation of permutations

[cloud walrus]

Spamakin🎷

[tough raven]

I suspect both of those are not simultaneously possible unless C is "one-dimensional" (maybe rank-1 projective over a commutative ring would do).

[tough raven]

AFAIK the terminology actually comes from considering bilinear pairings over non-fields (for ℤ at least), where injective ⇏ bijective.

E.g. <x, y> := x1 y1 + 2 x2 y2 + x1 y2 + x2 y1 between ℤ^2 and ℤ^2 (or simpler, <x, y> := 2xy between ℤ and ℤ) are non-degenerate but A → B* and B → A* have images of index 2.

[tropic obsidian]

yeah I saw it in the context of poincare duality giving you a perfect pairing on cohomology if you don't care about torsion

so I suppose in that case it's stronger than just non degenerate

[tough raven]

Yep.

The other two contexts I know of are lattices in number fields (you can do finite separable extensions of Dedekind domains if you want to be general) with the trace pairing, and root data (which I think are something like "root systems over ℤ" or root systems with a lattice between the root and weight lattices).

[neat zodiac]

that is kinda neat

[barren sierra]

it really is quite straight forward to prove

just gotta keep track of some notation

I would recommend you prove it

[crystal vale]

I want to show that in a commutative Boolean ring with unity, every ideal is principal.

I know if the ideal is finitely generated then it is.

But How can I do for arbitrary ideal I?

Hint?

[thorn jay]

That would imply every boolean ring is Noetherian, which isnt true, as any infinite product of rings isnt Noetherian

Do you perhaps mean finitely generated Boolean rings?

[crystal vale]

no for finitely Boolean ring, i know how to show it

[thorn jay]

In general its not true

Consider the boolean ring 2^N where 2 denotes the two element boolean ring and N is the natural numbers

Then you have an infinite ascending chain of ideals corresponding to powers of the surjective endomorphism
(x_1, x_2, ...) -> (x_2, x_3, ...)

[crystal vale]

i see

[thorn jay]

If every ideal was principal, then 2^X must satisfy the ACC

[thorn jay]

Which this disproves

[boreal inlet]

This is essentially proving that Conjugacy classes of S_n are determined by cycle types, and it's exactly what we need

[boreal inlet]

If your boolean ring is Noetherian, then you already have that it's a PID. Otherwise I don't think this is true

It's also Artinian, and also it must be finitely generated as a whole

All these are equivalent for Boolean rings

But if you have not finitely generated, as empeace said, it doesn't hold

[thorn jay]

Finite, yes

Boolean algebras are locally small

[ivory ore]

how to eliviate the frattini argument to be "somehow" applicable to infinte groups or in general groups

Let $P \le G$ be a Sylow $p$-subgroup. If $N_G(P) \le H \le G$, then $H$ is equal to its own normalizer; that is, $H = N_G(H)$.

[cloud walrus]

longboard kayak

[ivory ore]

also for naive reading now the question arises where in the case of infinite groups we can't argue the damn frattini's

or it's safe to assume we don't care about infinite cases at all(repeatedly mentioned in the text)

[rocky cloak]

Let G = <s, r: s^2 = 1, srs = r^-1> be the infinite dihedral group.

Then P = <s> is a 2-Sylow subgroup and N(P) = P.

Take H = <s, r^2>, then H is normal N(H) = G.

The problem is that G (and H) doesn't act transitively on its sylow subgroups. Is this what you call the Frattini argument?

[ivory ore]

idk, i was trying to find out what goes wrong in the infinite case

there was an exercise:

(i) Let $X$ be a finite $G$-set, and let $H \le G$ act transitively on $X$. Then $G = HG_x$ for each $x \in X$.\\

(ii) Show that the Frattini argument follows from (i).

[cloud walrus]

longboard kayak

[ivory ore]

so yeah, the transitive action is clicking the thing?

[ivory ore]

this seemed like a nice application to the argument, but there is no restriction of being finite

to which i found this which feels like the proof of frattini's argument but some how the constraint is not there!

[rocky cloak]

I mean it just pulls the P and xPx^-1 are H conjugate out of nowhere with no justification

[ivory ore]

oh yeah

uses the last lines of the argument

[candid patrol]

Hi Jagr

Remember about the problem : |Aut(G)| = 3, find G

There is no solution if G is finite

But what about if G is infinite ?

[ivory ore]

naively i have to say we can get some restriction on INN

as G/Z can't be no trivially cyclic and the only possibility here is 1 or Z3

so abelians?

[thorn jay]

Yeah Inn must be trivial

then every element must be of order 2, again, as else you'd have the nontrivial involution
x -> -x
Of order 2

I.e. we've got a vector space over F_2

[ivory ore]

from two non-trivial Aut elements right?

[thorn jay]

Well x -> -x is an automorphism of order 2, which is impossible by Langrange if |Aut(G)| = 3

So x = -x for all x

Thus G must be an F_2-vector space

There is probably an argument without choice, but with choice G has a basis B so you have at least Sym(B) automorphisms.
If |B| > 2 then |Sym(B)| > 3 which is impossible
If |B| = 0, 1, 2 then G must be finite which was already disproven

[glad osprey]

What is the addition of the vector space in this case? g + h := gh? Doesn't this require G to be abelian?

[ivory ore]

#groups-rings-fields message

[glad osprey]

ah, I see

I should learn to read

[rocky cloak]

The argument doesnt really have anything to do with the finiteness of G

.

[thorn jay]

I love assuming vector spaces have a basis

[rocky cloak]

Hey, if you can't find a basis that's a skill issue. No need to blame the vector space

[thorn jay]

If the argument works for any choice of basis then its a good enough argument for me

🔥

[tardy hedge]

excuse me??

[thorn jay]

Time travel

[mental silo]

who's theorem????

[tardy hedge]

Kummer

Number guy

Galois kinda guy too

[mental silo]

Freaky name

[tardy hedge]

Freaky 😛

😜

[mental silo]

(I have the maturity of a 13 year old)

[tardy hedge]

Me too bro

[keen badge]

What is the meaning of the notation in the middle(I guess it's model theory related, but I literally have no idea)?

[south patrol]

indeed there are various other typos too lol

"Is called on inner automorphism"

":-"

ye lol

[keen badge]

Lol

This is the only file I found on group automorphisms let alone inner automorphisms so this will have to do.

Maybe some of you have a pdf about group automorphisms and characteristic subgroup?

[keen badge]

I found 3 PDF's and 1 of them is not helpful(at all) - also, they are too short.

If someone knows a book that covers automorphisms, and characteristic subgroup I'd appreciate it.

[glad osprey]

Maybe Finite Groups by Gorenstein? https://bookstore.ams.org/view?ProductCode=CHEL/301

[thorn jay]

LMAO

T_a(x) = axa^-t does not deduce x in G

Would be the interpretation of

That

Left side is true, by definition so "T does not deduce x in G" can only mean that x is not in G

[languid trellis]

what text is this?

[tardy hedge]

Just my prof’s course notes

[keen badge]

https://magadhmahilacollege.org/wp-content/uploads/2020/09/INNER-Automorphism.pdf

[keen badge]

This is not a good one.
Maybe something else?

I can't find a PDF of it, and am not going to spend 70$ for one topic

[glad osprey]

I'm assuming you have looked through all the standard algebra texts, like D&F, Artin, etc?

[keen badge]

Don't know what D&F is

[glad osprey]

Dummit and foote

You can look through #book-recommendations , for algebra in general there's Lang, Hungerford, Aluffi (x2), and lots more I'm forgetting now

And for group theory in particular there are also loads, for example Actions of Groups or Finite Group Theory by Isaacs
https://www.cambridge.org/highereducation/books/actions-of-groups/9E2CF39D99D882E664FFAB8547B39469
https://bookstore.ams.org/view?ProductCode=GSM/92

[glad osprey]

@keen badge did you find what you were looking for? If not, I found a pdf you could use, but you have to DM me for it
-# 🏴‍☠️

[neat zodiac]

can someone explan c to me please

[wraith cargo]

Take x in R

0 in Z/2Z acts on R by the identity

1 in Z/2Z acts on R by sending x to -x

[glad osprey]

What does e) mean? Z/2 x Z/2 acts by reflection, ie sending edges to its opposite side?

[vestal girder]

it acts by reflections through the diagonals, and by sending edges to their opposites in pairs

if you have a rectangle with edges a,b,c,d it can act by (a b)(c d), (a c)(b d), (a d)(b c) and e

[neat zodiac]

How does that work, what do you mean by act?

[glad osprey]

through the diagonal? but that doesn't send the rectangle to itself

[glad osprey]

are you familiar with group actions?

[neat zodiac]

this is the beginning of the chapter of my notes

[neat zodiac]

it doesn't really explain it

[glad osprey]

interesting choice I would probably define group actions before I gave a bunch of examples

[vestal girder]

oh yeah oops im thinking of a way it acts on squares

yeah ur right

[neat zodiac]

oh literally under it it says this

[vestal girder]

text books dont have to be read in order, you can skip ahead to the definition and come back later

oh lol

[glad osprey]

I think it's just reflection through the other axis, just weird that they just say "by permutations"

[vestal girder]

i remember hearing some quote recently about how very few text books are meant to be read in order

[vestal girder]

yeah idk. if they had said pairs of disjoint permutations it might have been better

like every group action is a permutation in some way, so its kinda vague to say "by permutations"

[glad osprey]

yeah, agree

[glad osprey]

Hopefully the examples make more sense after seeing the definition. You can think of group actions in many ways, like a collection of functions as in the notes, or equivalently as a kind of "multiplication" f : (G, X) -> X, or even as a group homomorphism from G to the automorphism group Aut(X) (which is just the group of bijections on X if X is a set)

Note that if G is Z_2 then the possible group actions on X is pretty restrictive: f_0 must be the identity by definition, and f_1 must satisfy f_1 o f_1 = Id, since 1 + 1 = 0

[tardy hedge]

Group actions are always cool

We love puttin this shit into MOTION

[hidden wind]

it’s true

[amber burrow]

i just had a high school math comp

and there was one question

f: {1,2,3,4,5,6,7} -> {2,4,6,8,10,12,14} is surjective, and 1/2f(1/2f(1/2(f(n))) =n

and my sol was reminding me of group action stuff

but like not really related to groups at all

more so just permutations

but i guess thats just what a group action is

idk you can think of f as permuting {1,2,3,4,5,6,7}, since n |-> f(n) |-> 1/2f(n) which is still in the set. then its obvious to see either you have 2 3-cycles, 1 3-cycle, or none, with the rest mapping to themselves

i messed up the sol tho cuz i thought number of 3-cycles is 765/3! instead of 765/3

[violet spade]

Hi, I'm trying to prove the following statement:
"Let $K$ be a field and $f \in K[X]$ be an irreducible seperable polynomial of degree 5. Then $Gal(f)$ contains an element $\sigma$ of order 5". How do we prove this statement? I know that the action of $Gal(f)$ on the zeroes of $f$ is transitive because $f$ is irreducible, so for any $\alpha$ such that $f(\alpha)=0$, we have $\exists \sigma$ such that $\sigma(\alpha)=\beta$. But how am i supposed to go further with this information?

[cloud walrus]

joel

[violet spade]

Or is it actually really obvious?

I have a theorem in my book stating that this

And using Cauchy's theorem, we are done?

[sturdy spear]

I love algebra

[rocky cloak]

Yup that's it

[sturdy spear]

we proved that every subgorup of Z is of then form bZ

why this proposition

i dont see any prupose of this

it seems like artin is trying to define GCD using the above proposition

also this one

[crystal vale]

So now you can say Za + Zb is a cyclic subgroup

[sturdy spear]

but doesn't he state this here

(this is the paragraph just above this proposition)

[kind temple]

it characterizes the addition/intersection of subgroups of Z and is also a way to define the gcd/lcm

[sturdy spear]

interesting

using subgroups of Z to define gcd/lcm

[kind temple]

the converse of this also holds:

if you have dZ = aZ + bZ, then

1. d divides a and b
2. if e divides a and b, then e divides d
3. d = ra + sb for some r,s in Z

so d must be the gcd.

3. is kind of an extra property about the gcd that arises from this equality. 1) and 2) are the only ones which are necessary to pin down the gcd.

[crystal vale]

since Za + Zb = gcd(a,b)Z, so it is cyclic

[sturdy spear]

i see well its good time to prove this bi conditional thing

[sturdy spear]

i see,

[crystal vale]

if i have to show F[X]/<X^2 - 1 > is isomorphic to F[X]/<X^2> if char F = 2, where F is field.

Let char F = 2, then X^2 - 1 = (X-1)^2, so map F[X]/<(X-1)^2> -> F[X]/<X^2> by f(x) + <(x-1)^2> -> f(x+1) + <x^2>, then it is well defined homomorphism, and we can verify it is bijective, so it is isomorphic.

is it correct?

[velvet hull]

looks right
basic idea is that f(x) = ax+b is always an automorphism of a polynomial ring as long as a is a unit
an in fact all possible automorphisms of R[x] that fix the base ring must be of this form

[glad osprey]

Really basic question: $N = \begin{pmatrix}1 & 0 & b \ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}$ is a subgroup of $G = \begin{pmatrix} 1 & a & b \ 0 & 1 & c \ 0 & 0 & 1\end{pmatrix}$, but I'm struggling to work out what $G/N$ is. I think it's $\begin{pmatrix} 1 & a & 0 \ 0 & 1 & c \ 0 & 0 & 1\end{pmatrix}$, but this is not closed under multiplication as far as I can see. What am I missing?

[cloud walrus]

sheddow

[velvet hull]

not sure if there's a matrix representation of the group available in that form, but it should just be Z^2 no?

[glad osprey]

Yeah, or R^2 if the matrices are over R I guess

[velvet hull]

it's isomorphic to [1 a b; 0 1 0; 0 0 1] as a multiplicative matrix group but I don't see a way to derive this form ground up

[glad osprey]

I'm pretty sure the quotient is $\begin{pmatrix}1 & 0 & a \ 0 & 1 & b \ 0 & 0 & 1\end{pmatrix}$ btw, I made an error to begin with because the homomorphism I wrote down wasn't a homomorphism

[cloud walrus]

sheddow

[glad osprey]

yeah, or what you wrote down is correct too I think

you can use the homomorphism phi([1 a b; 0 1 c; 0 0 1]) = [1 0 a; 0 1 0; 0 0 1] for example

[cyan ice]

So let
[
G := \mathbb{U}_3(F) = \left{ \begin{pmatrix} 1 & a & b \ 0 & 1 & c \ 0 & 0 & 1 \end{pmatrix} \middle| a, b, c \in F \right}.
]
First one should convince oneself that
[
Z(G) = \left{ \begin{pmatrix} 1 & 0 & b \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \middle| b \in F \right}
]
so that since ( Z(G) \leq G ) is a normal subgroup, the quotient is well-defined.

Then we have that
[
gZ(G) = g'Z(G) \Leftrightarrow gg'^{-1} \in Z(G),
]
i.e., ( gg'^{-1} ) is of the form
[
\begin{pmatrix} 1 & 0 & b \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}.
]

But to find when it is the case that two matrices are related in this way, we first need to see if there is a general form for the inverse ( g^{-1} ) for ( g \in G ). I claim there is. By Gauss-Jordan elimination (or direct computation), we find that the inverse of
[
g = \begin{pmatrix} 1 & a & b \ 0 & 1 & c \ 0 & 0 & 1 \end{pmatrix}
]
is
[
g^{-1} = \begin{pmatrix} 1 & -a & ac - b \ 0 & 1 & -c \ 0 & 0 & 1 \end{pmatrix}.
]

Therefore, we see that
\begin{align}
gZ(G) &= g'Z(G)
\Leftrightarrow gg'^{-1} &\in Z(G)
\Leftrightarrow
\begin{pmatrix} 1 & a & b \ 0 & 1 & c \ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & -a' & a'c' - b' \ 0 & 1 & -c' \ 0 & 0 & 1 \end{pmatrix}
&= \begin{pmatrix} 1 & 0 & \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}.
\end{align*}

One finds that this holds whenever ( a = a' ) and ( c = c' ). Therefore, since the quotient does not discriminate on ( b ), we see that every element ( g \in G ) when viewed inside the quotient is equivalent to some element of the form
[
g = \begin{pmatrix} 1 & a & 0 \ 0 & 1 & c \ 0 & 0 & 1 \end{pmatrix},
]
so that each element in ( G/Z(G) ) can be represented this way. Therefore, it is indeed the case that
[
G/Z(G) = \left{ \begin{pmatrix} 1 & a & 0 \ 0 & 1 & c \ 0 & 0 & 1 \end{pmatrix} \middle| a, c \in F \right}.
]

I hope this is right (you can check the details!) 🙂

[cloud walrus]

Benjamin
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cyan ice]

ouch, that did not work as I wanted it (Ig this latex bot is different from the one I am used to)

hmm

Let me try this:

So let $G := \mathbb{U}_3(F) = \left{\begin{pmatrix} 1 & a & b\ 0 & 1 & c\ 0 & 0 & 1 \end{pmatrix} \middle| a,b,c \in F\right}$. First one should convince oneself that $Z(G) =\left{\begin{pmatrix} 1 & 0 & b\ 0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix} \middle| b \in F\right}$ so that since $Z(G) \leq G$ is a normal subgroup the quotient is well-defined.

Then we have that $gN = g'N \Leftrightarrow gg'^{-1}N = N$ i.e. $gg'$ is on the form $$\begin{pmatrix} 1 & 0 & b\ 0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix}.$$

But to find when it is the case that two matrices are related in this way, we first need to see if there is a general form for inverse $g^{-1}$ for $g \in G$. I claim there is. By Gauss-jordan elimination we find (I claim) that the inverse of $$g = \begin{pmatrix} 1 & a & b\ 0 & 1 & c \ 0 & 0 & 1\end{pmatrix}$$ is $g^{-1} = \begin{pmatrix} 1 & -a & ca-b\ 0 & 1 & -c\ 0 & 0 & 1 \end{pmatrix}.$$

Therefore, we see that
\begin{align*}
gN &= g'N\
\Leftrightarrow gg'^{-1}N &= N\
\Leftrightarrow \begin{pmatrix}1 & a & b\ 0 & 1 & c\0 & 0 & 1\end{pmatrix} \begin{pmatrix} 1 & a' & b'\ 0 & 1 & c'\0 & 0 & 1\end{pmatrix} &= \begin{pmatrix} 1 & 0 & \ast\ 0 & 1 & 0\ 0 & 0 & 1 \end{pmatrix}.
\end{align*}

One finds that this holds whenever $a = a'$ and $c = c'$. Therefore since the quotient does not discriminate on $b$ we see that every element in $g$ when viewed inside the quotient is equivalent to some element on the form $$g = \begin{pmatrix} 1 & a & 0\ 0 & 1 & c\ 0 & 0 & 1 \end{pmatrix}$$ so that each element in $G/N$ can be represented this way. Therefore it is indeed the case that $$G/N = \left{\begin{pmatrix} 1 & a & 0\ 0 & 1 & c\ 0 & 0 & 1\end{pmatrix} \middle| a,c \in F\right}.$$

[cloud walrus]

Benjamin
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crystal vale]

thank you

[ivory ore]

Show that $Q_8$ is isomorphic to a Sylow 2-subgroup of $SL(2, 5)$. (Hint. Show that $SL(2, 5)$ has a unique involution.)

[cloud walrus]

longboard kayak

[ivory ore]

well after knowing all the subgroups of Q8 one can tell what is the involution and where does it lie

however how to show it's the unique in the vector space too?

just by looking at the matrices ?

or take a general matrix and form a polynomial to do the characteristic analysis to then conclude from all the possible outcomes?

$SL(2,5)/Z(SL(2,5)) \cong A_5$.\\
$Z(SL(2,5)) = \{\pm I\}$, and $-I$ is the unique involution in $SL(2,5)$.\\
Since $A_5$ is simple and has no elements of order 2 lifting to more than one element in $SL(2,5)$, $-I$ is the only involution.

[cloud walrus]

longboard kayak

[ivory ore]

that sounds? tho i have shown no rigour in the claimed isomorphism!

https://groupprops.subwiki.org/wiki/PSL(2,5)_is_isomorphic_to_A5

gotta understand this proof now

i think we have to construct some action of the vectorspace (acting) on the plane

[ivory ore]

honestly i have no sophisticated knowledge on prjective line(which i would like to take care), but the action somehow seems intuitive here, it goes to S6 after reading a bit about the projective line for there are 6 elements and the kernel is C2

so factoring through the quotient we can show there is a injective mapping from the G/Z to S6

and now from the injection it's clear that it has to be A5? or something more?

so we standing on the fact: the only group of order 60 that can be embedded in S6 is A5

[warm badge]

how do i show b and c

[crystal vale]

for c, show every finitely generated subgroup is cyclic, and since the group is not cyclic, so it cannot be finitely generated

[warm badge]

how did u recognise that they will be cyclic

[velvet hull]

think about the lcm of the orders

[warm badge]

i only know that the orders divide p^m for some m

[velvet hull]

suppose G = (z1,z2) with the order of them being p^2 and p^3, whats the group generated by both?

[cyan ice]

b) If proper there is some element z not in H and z is the primitive p^nth root of unity for some mu_p^n but then it geneates this group and this mu_p^n contain all the smaller groups mu_p^k for k leqslant n. But this forces the group to only contain elements zeta such that zeta is a primitive root of unity for mu_p^ell for ell less than n since if larger then the larger contains the smaller so in particular contains the element z we excluded. So the group must be finite and every finite subgroup of the multiplicative group of a field is cyclic. 🙂

(I believe this work, sorry for not writing nicer).

[warm badge]

order divides p³?? btw isn't this argument simple enough that since all elements have finite order and the group is abelian, the order of a finitely generated set is bounded, but the mothergroup is infinite.

[velvet hull]

that's the sketch yes, you only need to show that the group generated by two elements is cyclic

and the argument can be modified to prove b)

[warm badge]

is that so man? if p³ divides the order then p³ ≤ the order. also (z1z2)^p³ = 1 this must mean the order is p³. is that right

[velvet hull]

the group is abelian, so (z1z2)^(p^3) = z1^(p^3)z2^(p^3) = z1^(p^3). I don't know how you're getting that that is equal to 1

in fact you can explicitly compute what the order of z1z2 is, but don't do that just yet. think about it conceptually. if you can't then do that

[warm badge]

aaah like i thought z1^p³ = (z1^p²)^p = 1, but if that's wrong i dunno

[velvet hull]

oh, wait, that looks right, I misread the original question since we're dealing with powers of p only that makes the question much easier

[warm badge]

ooh even easier what is the cyclic argument

[velvet hull]

hmmm, how about this one instead, it should be much cleaner (I'm assuming the FT of algebra for this one):

• ||how many solutions can the equation x^{p^n}-1=0 have in the complex plane at most?||
• ||Suppose G is generated by <z1, ..., zn>. Let p^n be the order of z1, and I also tell you in fact z1 has the biggest order among all the generating elements||. ||What is the size of G at least?||
• ||what is the size of G at most? What do you conclude about G?||

[warm badge]

first, n solutions ig. secondlyy aaaah is that gonna be p^n??

[velvet hull]

p^n solutions

[warm badge]

oh yeah p^n solutions

size of G is atleast p^n?

at most i dunno, well atmost we gotta consider the order of all other elements and multiply them with p^n

[velvet hull]

what is z2^(p^n)

[warm badge]

are you saying it's gonna be 1?

[velvet hull]

why?

[warm badge]

then what else... i can't tell anything about z2^(p^n)

[velvet hull]

that's all you need.

so if g is some random element in G, what is g^(p^n)

[warm badge]

if g is a random element, then g^(p^n) = 1 i think

[velvet hull]

can you tell me why?

[warm badge]

g is a word of z1,z2,...,zn and z1^(p^n) = 1, thuus

[velvet hull]

no, that's not enough

[warm badge]

yes right we should think about the rest of the elements

p^n is the largest order you said right?

oh then it is indeed 1, all other element has order less than p^n

[velvet hull]

that's still not enough, one keyword I'm looking for

[warm badge]

lcm?

[velvet hull]

abelian

Q. if the order of f is n and the order of g is m, what do I know about the order of fg?
A. fucking nothing, if you don't tell me that fg=gf

[warm badge]

true true

[velvet hull]

and in fact for every integers n,m,p there exists a group where the order of f is n, the order of g is m, and the order of fg is p

[velvet hull]

but now go back to looking at this

[warm badge]

size of G is at most p^n ??

man i see

G is cyclic

[velvet hull]

so this is one way to prove it, it does require you to assume some things about polynomials in C but it's a really slick proof

your class may or may not be okay with those assumptions

[warm badge]

hmm but well they expect us to have read complex analysis, which I didn't anyways, but i see the proof and this can surely be modified to prove b... or can it? does that mean proving every proper subgroups are finitely generated

[velvet hull]

oh, the question's not assuming finitely generated?
that's not too bad of a fix, you just need to show that subgroups of Z(p^infty) are downwards-closed

i.e. if G is a proper subgroup with an element g of order p^n, then every element of order less than or equal to p^n are also in G (the proof we just did does this)

and now think about why that means G has to be finitely generated

[warm badge]

that can be shown indeed but how do i show elements of order more than p^n are not in G? from the question, it seems there must exist some n, for which all elements z s.t. z^(p^m) = 1, m>n, z is not in G. so i gotta show the existence of such n first. this should be true since if there is no upper bound of such n then G doesn't remain proper. but the other part?

[velvet hull]

you don't have to

[warm badge]

ohh why is thaat

[velvet hull]

everything smaller than G is in G, and there are only finitely many such elements

so what happens if G is not finitely generated?

[warm badge]

G takes on the whole mothergroup

[velvet hull]

you skipped a few steps there, that's not necessarily true

it's because Z(p) is countably infinite

or in human words, there are no "in between" infinities

[warm badge]

Z(p) is the set of all z with z^p = 1 and there are at most p solutions, so isn't it finite!!

[warm badge]

i can only tell then G is not finite then

[velvet hull]

that is true, but I can keep increasing the p

so z^p=1, then z^(p^2)=1, z^(p^3)=1 etc

this is a very infinite group

[warm badge]

hmmm that's right

[velvet hull]

so here's a hint - subgroups of Z(p) are characterized by the biggest number n such that z^(p^n) have a solution in the subgroup

[warm badge]

so to begin with the proof i take a proper subgroup G, if there is no maximum n for which z in G has z^(p^n) = 1, then G is simply the mothergroup by definition. now we only prove the size of G is p^n and we're done, and we're gonna do it kinda the same way we proved c ig.

[velvet hull]

sounds like a proof

[warm badge]

yehhh thankss dude, have been hanging on it since the morning

[chilly ocean]

Hello, guys I am trying to learn proof of unsolvability of quntic polynomial using galois theory.

This is what i have understood till now. There is a one to one relation between Field extenstion and Galoi group of that field extension. If the field is extended by use of radical then it's galois group is abelian and the base field is abliean and normal group to it. This is what i think is going on. But I am not able to understand why that is the case

[coral spindle]

I would suggest getting a book that covers Galois theory. Perhaps Fraleigh.

[chilly ocean]

Thank you for your recommendation!

[cloud walrus]

🇵🇱 uuybnuuy

[thorn jay]

G is just the real numbers though

[delicate orchid]

I think the idea here was to just show it's iso to R

[coral spindle]

G = R. They are literally equal.

So anything you say about the set G is also true about the set R.

A rose by any other name

[delicate orchid]

but if we must - yes, x^3+y^3 is ((x^3+y^3)^(1/3))^3 so it's closed. It's borderline tautological

yeah I remember (all those years ago 👴) that at these early stages the things you're showing are so obvious they become difficult

[thorn jay]

This seems like a trick ish question lol

[coral spindle]

Or a question to check if you're paying attention

[thorn jay]

I always get anxious that im missing something, no matter how well I know the subject matter

[fallow zephyr]

Hello
I'm working through a set of elementary proofs for groups, can I post it here to get it verified?

[tardy hedge]

yes

[fallow zephyr]

gimme one sec

[tardy hedge]

ok

[fallow zephyr]

This is the problem set

proof for number (a)

I use proof by contradiction and I wanna know if this is how it's intended to be done

proof for numbers (b) & (c)

[thorn jay]

Next time just post it, its what this channel is for :)

[thorn jay]

This looks good

[thorn jay]

This isnt contradiction, its direct

[fallow zephyr]

O
but I'm assuming b and c are distinct right?

[thorn jay]

No

Youre just assuming that they satisfy the same property, which implies they are the same

Thats a direct proof of uniqueness

[fallow zephyr]

I see

thank you!

[glad osprey]

"not necessarily abelian"... uses additive notation but with oplus instead of + 💀

[thorn jay]

Thank you I was going crazy too

EVEN WORSE

OPLUS

WHY

Why are we treating them like tensors

Istg

[glad osprey]

yeah, like let's use oplus to prevent any confusion with ordinary addition, but let's also call the identity element 0, because why the heck not

[fallow zephyr]

yeah true my bad

[thorn jay]

IN THIS HOUSEHOLD WE USE 1 AND •

[fallow zephyr]

I wanted to quickly write it down, I'll pay attention next time

[glad osprey]

but you didn't write the exercise sheet I'm guessing? We're not making fun of you

[thorn jay]

Were making fun of the mf who did, yee ass convention

I saw someone use W(X) for a general free algebraic structure
!???
W?????

[glad osprey]

haha, what a loser
-# I don't know what the correct convention is

[thorn jay]

F_K(X)

F because Free algebra

And K because its always with respect to some class of algebras

[glad osprey]

ah yeah, I've seen that, atleast for groups

[thorn jay]

Yeee

There the K isnt there, but in my notation I'd use Grp for the class of groups

[ivory ore]

my eyes

[thorn jay]

Abelian group G^-1 such that G \oplus G^-1 ≈ 0

[fallow zephyr]

am I correct in understanding here that the proof here requires me to show that there is no common element at all, and that's all I'll ever need to demonstrate?

[thorn jay]

No, you still have to prove that theyre the same if g - h is in S

Also, "S" for subgroup qwq

[fallow zephyr]

I see

I'm finding the proof a bit confusing as I don't immediately see the relation between the difference of g and h and the cosets being disjoint

[thorn jay]

Suppose that S + g and S + h had a common element, say x. This means that x = a+g and x = b+h for a, b in S
=> a+g = b+h
=> g-h = -a + b
=> g-h in S

You can fill in the steps where needed

[fallow zephyr]

a & b would be present in S already right?

[thorn jay]

Yes

"For a, b in S"

[fallow zephyr]

I see, that makes sense

I was kinda tripping up over that

cause I just assumed s & s' don't necessarily have to be in S

[thorn jay]

Its by definition of coset

[fallow zephyr]

yeah I think I haven't fully understood it

thanks for the pointers

[thorn jay]

👍

[delicate orchid]

Words in X

[thorn jay]

oh that makes sense

I still dislike it though

[delicate orchid]

I do too

[thorn jay]

oh and also "G" for a general algebraic structure

which is just, a crime

[delicate orchid]

Is this paper from the 1960s or something what is this

[thorn jay]

2002

https://arxiv.org/abs/math/0204245

[dark egret]

Hi, is anyone available for help to understand a proof from J.S. Milne's book : Fields and Gallois Theory?

[next obsidian]

[dark egret]

Here is the proof i'm having trouble with ^^.

I have another way to prove a) but the proof from the book makes no sense to me. And I feel like I might be missing something important.

My proof is quite simple,
as any element of F(a) is in the form P(a) where P is in F[X], this means that for every phi (homomorphism), phi(P(alpha)) = P(phi(alpha)). From there it's pretty obvious that chosing phi(a) gives the result. Also, alpha can only be transcendantal because it would be absurde do to phi(P) = P(phi)

What makes no sense to me :

• The first sentence : I read F[a] is isomorphic to F[a]. I don't know what he's trying to say there
• the second sentence feels so wrong, I think I'm missing something. gamma = 1 cannot be chosen as an image of alpha through phi as it would mean that phi is not injective but it is because it is a homomorphism of fields...
• same goes for the third sentence. If I understand corectly, there can't be nonzero elements mapping to zero as phi is injective

[next obsidian]

F[alpha] means the F-subalgebra of F(alpha) generated by alpha. It’s the polynomials in alpha, which because alpha is transcendental means it’s isomorphic to F[x]. There’s a map F[x] -> F[alpha] sending x to alpha, this being injective is saying alpha is transcendental

Now because F[alpha] is a polynomial ring, you can send alpha to any element of Omega to define a ring map. You can even send it to 1, this won’t be injective but this is okay because F[alpha] is a polynomial ring not the field F(alpha). But, F(alpha) is the field of fractions of F[alpha].

Now the point is, for what choices of the image of alpha do you actually induce a map from F(alpha) well you need the map to send nonzero things to units so that p/q can map to phi(p)/phi(q). but as Omega is a field, this is just saying phi is injective

So when is the choice of omega in Omega going to make the map F[alpha] -> Omega by p(alpha) -> p(gamma) injective? Precisely when gamma is the root of no polynomial with coefficients in F, aka when gamma is transcendental

[dark egret]

I'm re-reading it all, but it already makes a lot more sense to me, thanks!!!

I was definitely confused with F[X] and F(X)

I thought that F[alpha] = F(alpha) but that is only true when alpha is transcendental

It's all clear, thank you for your time and explanation

[next obsidian]

No, not true when alpha is transcendental

This is like saying rational function F(X) = Polynomials F[X]

[dark egret]

yeah, that's what I meant 😆

[next obsidian]

Ah okay

[dark egret]

but it is when alpha is algebraic if I understand correctly

[next obsidian]

This is true

And maybe a proof is like so, F[alpha] = F[x]/I

Where I is the kernel of F[x] -> F(alpha) where you map x to alpha

Errr

Okay sure

When alpha is algebraic I is nonzero, so its principal

And so it’s F[x]/(f(x))

If f(x) wasn’t irreducible, by the CRT you get that F[x]/(f(x)) is a product of rings, but then it isn’t an integral domain so it can’t be equal to F[alpha]

So we get that F[alpha] is a field

And so any nonzero p(alpha) has an inverse, and so you can write all ratios p(alpha)/q(alpha)

[next obsidian]

Addendum, if f(x) = p(x)^n where p is prime, then you can’t use CRT, but this isn’t an integral domain all the same

[dark egret]

p(x)^n wouldn't be irreudible then

[next obsidian]

Yeah

[dark egret]

The proof I can think of is with the minimal polynomial of alpha over F which can't have 0 as a root. Thus the constant coefficient is non zero and we can factor the rest of it by alpha, giving an explicit form of the inverse (multiplying by the inverse of the constant coefficient)

[next obsidian]

I am trying to show that if f(x) needs to be irreducible

Because otherwise F[x]/(f(x)) isn’t an integral domain

But actually this follows because quotient is an integral domain if and only the ideal is prime

[next obsidian]

Yeah I think you can just do it manually too

I was thinking of doing that but wanted to handle it via like maps from polynomial rings, since that’s the flavor of the proof you had shown

[dark egret]

Yeah, I understand. That's also the reason I wanted to reed this kind of books. I always find abstract solutions to problems that I never find or not with such elegant solutions. I'm trying to learn how to do more mapping, factoring functions with quotients, stuff like that

I always feel stupid when writing down 2 pages and then someone draws 3 arrows and calls it a day

[next obsidian]

Hahaha

[opaque ridge]

hi

[worthy solar]

when you take a direct product or direct sum of rings / groups what do you call the individual rings/groups that make up the sum / product?

like

Z/3Z x Z/4Z

you would say "Z/3Z" is a _____ of the direct product Z/3Z x Z/4Z

like does it have a name

[wraith cargo]

direct summand

[thorn jay]

thats for direct sums

for products I think its just "component"

or factor maybe?

[worthy solar]

and like for the elements of the direct product what do you call the individual entries in the direct product?

like (r1, r2)

[thorn jay]

components? entries?

[worthy solar]

okay

i guess "component" or "element" wise operation then makes sense

[worthy solar]

Yeah that is what i was thinking

[worthy solar]

TIL

[thorn jay]

that's only when referring to direct sumsthough

though*

[worthy solar]

yeah i get that

[glad osprey]

I want to show that a simple solvable group is cyclic. I'm kinda stuck, I can't think of a way to show that a group is cyclic other than finding a generator. Anyone have any hints?

nvm, figured it out. We must have G' = {e}, so G is abelian, and therefore has no subgroups, so it is cyclic of prime order

[thorn jay]

Yessir

[south patrol]

Note this works even for G infinite, since you can still conclude it's abelian and then just pick any element to show it's cyclic lol

And then rule out the only extra case which is Z

[glad osprey]

ah yeah, didn't think about that case, but I guess it works anyway

[warm ember]

is it true that the algebraic closure of F is the set of roots to all polys in F[x]

no right

i cant think of a counterexample

[glad osprey]

isn't that pretty much the definition of algebraic closure? the set of all elements that's a root of some polynomial in F[x]

[warm ember]

how do we know this set is always algebraically closed

[warm ember]

i always thought it was this

but all the proofs that algebraic closure exists doesnt say anything about htat

[glad osprey]

I was about to say that if there is a polynomial in F[x] with a root that is not in F-bar, then you could adjoin it to F-bar to get a non-trivial algebraic extension, but it's kinda weird to talk about roots outside the algebraic closure. I think it makes more sense to talk about F-bar as the splitting field of the set of all polynomials

[warm ember]

i meant what if theres a root of a poly in \overlien{F}[x] outside F bar

if F bar is defined as the splitting field of the set of all F polys

[glad osprey]

hmm, then you could adjoin that root to F-bar to get an algebraic extension, which is impossible since F-bar is algebraically closed

or rather, it's impossible to get a non-trivial extension

[warm ember]

thats assuming the splitting field of all F polys is algebraically closed

[glad osprey]

I think you can argue that the only irreducible polynomials in the splitting field are linear, therefore the only algebraic extensions are trivial

[warm ember]

o

[warm ember]

how come proofs that algebraic closure exists isnt just proving this

[glad osprey]

hmm I guess it's easy to prove that the splitting field of a finite set of polynomials exists, but maybe it's harder if the set is infinite?

[warm ember]

i think this is false

its true for R and Q

but in general false

[glad osprey]

why?

[warm ember]

i cant find anything that says otherwise

[glad osprey]

It's late at night, but I think this works:
Let F be a field, and E be the splitting field of all polynomials in F[x]. Assume there exists an algebraic extension E <= E', and let a in E'. Let f(x) be the minimal polynomial of a over E, and let g(x) be the minimal polynomial of a over F. Then g(x) splits in E, so g(x) = (x - a_1)(x - a_2)...(x - a_n), and we know f(x) divides g(x), so f(x) must be linear. Therefore E = E'

[crystal vale]

i am taking phi: K[x,y] -> K such that f(x,y) -> f(a,b). It is clear that < x-a, y-b> is in ker phi, how can i show other direction? hint?

but K[x,y] is not ED

<y> is prime ideal in k[x,y], right?

[velvet hull]

(not 100% sure on this proof, someone double check this for me please)
k[x,y]/<x-a,y-b> ~= k[x,y]/<x,y> = k

[thorn jay]

Seems like the kernel of the evaluation map k[x, y] -> k sending x to a and y to b, so yes seems good

[velvet hull]

well it is, and that's the proof idea OP had

but showing explicitly that the kernel of the ev map is generated by the linear poly's is annoying, so I instead opted to show that x->x-a, y->y-b are automorphisms, which is easier

[thorn jay]

Oh like that

Yeah no thats good

[crystal vale]

oh so K[x,y]/<x,y> -> K by f(x,y) -> f(0,0)

[thorn jay]

Ye

[crystal vale]

i got it, thanks HChan

[warm ember]

how do you know g exists

an algebraic extension of an algebraic extension is algebraic?

so the algebraic closure is the largest algebraic extension

anyway thanks for stuff

[ivory ore]

anyhint on this?

[rocky cloak]

Well, there's a pretty big hint there to consider S3xS3. So what are the 2-Sylow subgroups of S3xS3?

[ivory ore]

oh

so two are S3

which has nontrvial intersection

and i need to find something else

[velvet hull]

wait wait wait. how exactly is S3 a sylow subgroup of S3 x S3

[ivory ore]

Let $G = S_3 \times S_3$. Then $|G| = 36 = 2^2 \cdot 3^2$, so the Sylow $3$-subgroups have order $9$. In $S_3$, the $3$-Sylow subgroups are cyclic of order $3$—each generated by a $3$-cycle. \\


Let $A = \langle ((1\ 2\ 3), \text{id}) \rangle$, $B = \langle (\text{id}, (1\ 2\ 3)) \rangle$, and $C = \langle ((1\ 2\ 3), (1\ 2\ 3)) \rangle$.

[cloud walrus]

longboard kayak

[ivory ore]

any fault here?

[velvet hull]

a sylow 3-subgroup in S3 x S3 needs to have order 9, your A,B,C all have order 3

[rocky cloak]

There's also only one 3-sylow subgroup, so that is maybe not so interesting

I guess bonus exercise is that the sylow subgroups of GxG' is ||of the form PxP'||

[ivory ore]

oh yeah i think i proved that

lacking memory

[ivory ore]

I'm illiterate

[ivory ore]

maybe i need to take rest, can't think of anything for this rn

[cloud walrus]

longboard kayak

[ivory ore]

this sounds?

[glad osprey]

Yeah, a is algebraic over E, so it must be algebraic over F. You could also think of the algebraic closure as the smallest field where every polynomial splits, which is isomorphic to the splitting field by uniqueness

[amber burrow]

A transitive permutation group $G$ on a set $A$ is called doubly transitive if for any (hence
all) $a \in A$ the subgroup $G_a$ is transitive on the set $A \setminus {a}$

[cloud walrus]

donut123

[amber burrow]

I feel like im doing something wrong

Consider $b \in B \subset A \setminus {a}$

[cloud walrus]

donut123

[amber burrow]

Then

wait nvm

[crystal vale]

I don't know how to do this problem.

And in the b part, is it mean that S is finitely generated subring?

Any hint?

[amber burrow]

Can someone double check my proof that double transitive implies primitive

From a previous exercise, a group is primitive if and only if for all a in A (the set being acted on), G_a is a maximal subgroup

Consider a subgroup H containing G_a

Suppose H is not G_a, so it contains something else

then there is some $\sigma$ such that $\sigma \cdot a \neq a$

[cloud walrus]

donut123

[amber burrow]

since $G_a$ acts transitively on $A\setminus {a}$, $H$ acts transitively on $A$

[cloud walrus]

donut123

[amber burrow]

Thus, $|O_a| = |G:G_a| = |G:H||H:G_a| = |G:H||O_a| \implies |G:H| = 1 \implies G = H$

[cloud walrus]

donut123

[chilly ocean]

i think for (a) you can have a look newton divided difference interpolation formula all the coefficients of that polynomial arebinomial coefficients

[chilly ocean]

for (b) it is saying that it is finitely generated Z module

Show that the element $xy -zw$ is an irreducible element in $C[x, y, z, w]$.

[cloud walrus]

Curvature

[south patrol]

Consider it as an element of C[y,z,w][x], say

[chilly ocean]

so C[x,y,w,z]/<xy-zw> = C[y,z,w]/<zw>

[south patrol]

That's not rly how it works

It's just that this is a degree 1 polynomial in x

[chilly ocean]

ohke

[crystal vale]

Any hint for part b?

[chilly ocean]

how it will benefit me

[chilly ocean]

is it free module over Z?

[crystal vale]

I have no idea

[chilly ocean]

assume that it is finitely generated , so there must be some polynomial of degree with binomial coefficients of sufficiently lasrger degree

tell me is it correct or not, otherwise i will have to work out

[crystal vale]

Why binomial coefficients?

[chilly ocean]

then f(a) will be integers for every integers a

[chilly ocean]

are you following conrad notes on doubly transitive group action

[amber burrow]

No, dummit and Foote exercise

[ashen hill]

anyone have a copy of algebra by artin?

pdf?

[cinder fox]

Is this proof all good?

[rocky cloak]

You know the library is the genes**.is** of all knowledge.

[chilly ocean]

PLEASE have a look at my question

[grizzled crow]

Do all texts generally consider the identity permutation to be a cycle (of length 1)?

[amber maple]

does anyone have any book recommendations for learning galois theory, i have a background in field theory and have some mild pre-existing knowledge of galois theory

[chilly ocean]

lang is best

[grizzled crow]

Seems to vary

[chilly ocean]

$\frac{\mathbb{R[X,Y]}}{<Y-X^2>}$ is ufd?

[cloud walrus]

Curvature

[tall karma]

can anyone help me understand it now so like if a, b are subgroups of H then if a*b still belong to H idk

[untold torrent]

If G is abelian group then

AB=BA

(A.B)^n=A^n.B^n or B^nA^n or both?

[chilly ocean]

that is the subgroup requirement ,

[tall karma]

so non abelian is AB does not equal BA

[untold torrent]

..

Group is abelian means it will hold for all elements

[tall karma]

ok

[untold torrent]

Commutativity

[tall karma]

is there like a way to tell if a group is abelian or not

nvm

[grizzled crow]

Group elements never leave their group when you multiply then with each other.. thats a key requirement of being in a group (assuming the operation is multiplication)

[tall karma]

so if a and b belong to g ab still belongs to g?

[grizzled crow]

Yeah if g is a group

[velvet hull]

R[x,y] surjects onto R[x] with x -> x, y -> x^2, and by the 1st iso it is isomorphic to R[x,y]/<y-x^2>.

But R is UFD, so R[x] is UFD

[chilly ocean]

can anyone see this, and ST=TS implies S invariant subspace

[late matrix]

if v is a λ-eigenvalue, then S(v) is too, since TS(v) = ST(v) = λS(v)

and so $S(V) \subseteq V$ where $V$ is the eigenspace of $v$.

[cloud walrus]

ark

[crystal vale]

Since on Q, it makes Cauchy functional equation so if f: Q ->Q is automorphism then f(x) = xf(1), and since it is automorphism 1 maps to 1, hence f(x) = x.

And if g is a field automorphism on R then it must be field automorphism on Q.

Then g(x) = x for all x in Q.

But I think I have to show g is continuous at 0.

I know that k ≥ 0 iff g(k) ≥ 0.

How to proceed further?

[kind temple]

try to show that g is order preserving using the hint you have now.
then, use the fact that the rationals are dense in the reals, i.e., if g(x) != x, then there is a rational q between g(x) and x

no need for any more tools from topology/real analysis other than density

[crystal vale]

Yes g is increasing

[crystal vale]

I got it, thank you @kind temple ❤️

[pearl fog]

does a^b-1|a^c-1 iff b|c or does it work only for prime?

[south patrol]

What is the context?

Oh, do you mean $a^b - 1 \mid a^c - 1$

[cloud walrus]

Prismatic Potato

[south patrol]

I read it as a^(b-1) lol

[rocky cloak]

If b divides c, then a^b - 1 divides a^c - 1

That works for all a.

a=0 breaks the converse at least

a=-1 also breaks it

And a=1

[pearl fog]

eh, just wondering after spending entire afternoon and night staring at finite fields

[south patrol]

Dw yeah I just didn't udnerstand the typesetting originally but ye

[pearl fog]

that direction simply guaranteed by polynomial factoring, the converse is the problem

prime is probably right cuz irred in Z so it looks like x then by some corollary of gauss' lemma

no idea on composites

[rocky cloak]

If c = bq + r with r<b, then the remainder of dividing
a^c - 1 by a^b - 1 is a^r - 1, so assuming a>1 so that a^r - 1 > 0 you get your iff

https://math.stackexchange.com/a/128013/306319

[south patrol]

oh lol nice

[pearl fog]

hmm maybe i can make this a problem for my junior high tutor

[south patrol]

apparently this too lol

well, presumably just same as what jagr has said

[rocky cloak]

This actually gives the gcd though, so it's going the extra mile

But yeah, same idea

[south patrol]

Yeah that's why I think this is nice hehe

I got thrown off by the originL thing you shared cause I thought x meant the polynomial ting there lol

But x is still just an integer lol

[chilly ocean]

suppose we have a basic nil potent block X of order n what is the dimension of the space of matrices Commuting with X

[chilly ocean]

i want to learn to find galois group of cubic and quartric any friendly resources

[rocky cloak]

I'm assuming you mean X is an nxn matrix with nilpotency n.

In any case, you can think about the C[x] module defined by X. A matrix commuting with X is exactly an endomorphism of this module.

So in this case the module is C[x]/(x^n), so the space is n-dimensional (given by polynomials in X)

[chilly ocean]

do you mean all the space of commt=uting matrices is of dimension n

[rocky cloak]

It's an n-dimensional space yeah. Specifically spanned by
1, X, X^2, ...

[chilly ocean]

i did it for n=3 and i am getting answer 6, by explicit calculation

[rocky cloak]

X = [0, 1, 0; 0, 0, 1; 0, 0, 0]
A = [a, b, c; d, e, f; g, h, i]
AX = [0, a, b; 0, d, e; 0, g, h]
XA = [d, e, f; g, h, i; 0, 0, 0]
Which means
d=0, a=e, b=f, g=0, h=d, i=e, g=h=0
So
A = [a, b, c; 0, a, b; 0, 0, a]

That's 3 dimensional

[chilly ocean]

can you explain why endomorphism is helping us to find the dimension

[rocky cloak]

Well because I find it easier to compute an endomorphism ring than to multiply a bunch of matrices.

The endomorphism ring of R/I is just itself, and dim C[x]/(x^n) is n

Very easy

[chilly ocean]

that is my question , how every endomorphism is commuting matrix, i have never seen this trick

[rocky cloak]

Well, it's just the definition of homomorphism.

A homomorphism of C[x] modules is a linear map that commutes with the action of x

[chilly ocean]

okay makes sense

[chilly ocean]

can you show the endomorphism is itself

[rocky cloak]

Well it's cyclic, so determined by where 1 is mapped. And because the ring is commutative you can send it anywhere

[chilly ocean]

but here we have module not ring

[rocky cloak]

A module is defined over a ring

R/I is an R-module

If R is commutative (or just if I is twosided), then End_R(R/I) = R/I

[chilly ocean]

oh fine

but @rocky cloak i never thought of this question using module i thought i can be done in pure linear algebra

[rocky cloak]

I guess you can look at the flag of generalized eigenspaces, and then conclude that the matrix must have this layered upper triangular form

Or just write out the matrix multiplication with eistein notation or whatever

[chilly ocean]

but module method is every nice i think it shouldbe used whenever possible

[glad osprey]

If n = [F(a) : F] and k = [F(b) : F] are relatively prime, then [F(a, b) : F] = nk, right? Or more generally if n = [K : F] and k = [L : F] are relatively prime, then [KL : F] = nk?

[rocky cloak]

Yes, in general [KL:F] will be a multiple of the lcm

(and divide the product)

[amber maple]

can someone help me understand this induction idk why i just need it spelled out a little clearer

gamma(L/K) is the galois group

the case n = 0 is that L = K(alpha_{0}) right? which is just L = K and you get the trivial group {id}

i figure thats the only way the base case is trivial

the induction seems as though its heading towards G(L/L) tho, i really dont know why im not understanding it

[rocky cloak]

Your image seems to cut out right before the actual proof starts.

But anyway, the basic idea is that if K(a2, ..., an) is normal, then G(L/K) is an extension of G(L/K(a2, ..., an)) and G(K(a2, ..., an)/K).

The former is a group of prime order, hence abelian, and the latter is solvable by induction.

The tricky part is just making sure this is normal.

You can do this by making sure it contains the roots of unity by adjoining them first (or last I guess, according to the numbering)

[amber maple]

the rest of it just goes if a1 is not in K then... but that part is fine

[rocky cloak]

Then I'm confused which part you find difficult

The n=0 case is just L=K yes

[amber maple]

i think my brain just isnt working today

would the n = 1 case be L = K(a1)

or is the induction that we are just taking extensions of K, like K' = K(a1) and working with L = K'(a2,...an) until we have L = K*

[rocky cloak]

Yeah, n=1 would be L = K(a1)
n=2 would be L = K(a1, a2)
general n would be L = K(a1, ..., an)

[amber maple]

isnt the order of G(L/K(a2, ..., an)) just 1 since L = K(a2,...,an) so ur just doing G(L/L) which is just {id}

i dont get what u mean by then G(L/K) is an extension of G(L/K(a2, ..., an)) and G(K(a2, ..., an)/K) this book doesnt talk about groups being extensions of other groups

[rocky cloak]

The group in question is
G(L/K)

If L = K(a2, ..., an), then that's n-1 elements. So by were in the n-1 case

So by induction G(L/K) is solvable

[rocky cloak]

I was just describing my guess at what happens after your image cuts out

... But it seems you understood that part...

[amber maple]

maybe i need to work on this tomorrow i feel stupid today

galois theory is hard 😭

[rocky cloak]

Like you're doing induction on the number of as. If a1 is in K, then you can remove it and now you have fewer as

That's it

[amber maple]

wait

[rocky cloak]

There is no higher math or anything complicated going on.

It's just if you have n things and remove one you now have n-1 things

[amber maple]

i think im working now

yeah

no idk what i was thinking

[chilly ocean]

ㅡㅡㅡㅡㅡㅡㅡㅡㅡ
Why it holds

[velvet hull]

what is K? an arbitrary field?

[rocky cloak]

K\{0} is a group under multiplication with q-1 elements

[chilly ocean]

ㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡㅡ
Why this holds..

[coral spindle]

Did you understand why what jagr said up here is true

Or relevant, rather.

If you did, then this should be immediate.

If you didn't, now's the time to say.

[crystal vale]

Do I need R to be integral here?

Someone gave me this proof:

Let x in R, say R has dimension n, then 1, x,...,x^n, those vectors are linearly dependent.

So there exists c_i's in C, at least one c_i is non-zero such that c_0 + c_1x +..+c_nx_n = 0, so x is a root of non-zero polynomial in C[y], and C is algebraic closed so x in C.

Is it correct? But here we didn't use an integral part of R

Yes but see the proof, we didn't use the integral part of R

[velvet hull]

well the big assumption here is that there does exist such an x

if we know that R is a field, for instance, then we know that at least one x exists by the primitive field theorem

[velvet hull]

an immediate counterexample I can think of is the ring C^2

(0,1) * (1,0) = 0 so this is not an integral domain

[crystal vale]

But how does C^2 contain C ?

[velvet hull]

x -> (x,0)

[crystal vale]

Yes but I thought it means that C \subset R

[velvet hull]

well if you wanted to I'm sure you can modify the structure of R such that that is possible

I don't see the distinction between that and there just existing an embedding of C into R I sidestepped this issue, see below

[crystal vale]

I don't get if

[velvet hull]

R being an n-dimensional vector space over C just implies that there exists vectors v1, ... , vn such that they span R over C

this does not guarantee that the basis vectors are all the power of some single number

[crystal vale]

No, I pick x such that 1,x,x^2,..,x^n are n+1 vectors so they must be LD

[velvet hull]

no

you cannot assume that you can always find such an x

that's not the definition of a vector space

[crystal vale]

I just pick x

[velvet hull]

what if everything in R was nilpotent with degree less than n?

if R is not an integral domain I don't see why that isn't possible

[crystal vale]

Then x^n = x^m

[velvet hull]

no, x^m = 0 for some m < n

that's the definiton of nilpotency

not x^m = 1

[crystal vale]

Can you please listen first?

I am saying that take x in R

Then take vectors 1,x,x^2,..,x^n, if they are not distinct that means x^i = x^j, for some i≠j, then take polynomial y^i - y^j, so this polynomial has root x, so x in C.

If they are distinct vectors, since dimension is n, so this set of vectors must be LD

[velvet hull]

distinct does not imply LI, though

[crystal vale]

See dimension is n, so how can n+1 vectors are LI?

[velvet hull]

let me just find a counterexample.

I claim that you need integral domain, because you need R to be a field for this to work

[velvet hull]

C[x]/<x^2> is a ring containing C, and it is also a finite dimensional vector space over C but clearly non isomorphic

so, run your argument through me again and let's figure out where the problem is, if there is one, together

[crystal vale]

Yes, but then x^2 = 0 in C[x]/<x^2>, so y^2 = 0 has a root x, cannot I say x in C?

It makes no sense

[velvet hull]

I do not know what you are trying to do

[crystal vale]

Where am I wrong?

I mean x^2 = 0 in C[x]/< x^2>, right?

So that's means x satisfy y^2 = 0 equation

Can I use C is algebraic closed here?

[rocky cloak]

The quaternions is a finite dimensional noncommutative domain that contains C.

However, you don't need the full commutativity assumption, it's enough to assume C is in the center

[rocky cloak]

You're using that the c_i commute with x, for this to be a contradiction

[south patrol]

Corollary: M_2(C) is not a domain lol

[chilly ocean]

THE integral domain i think is used where x non zero implies x^n is nor zero

[crystal vale]

Okay say R is commutative then where I used R is integral?

[rocky cloak]

The integral part is just commutativity. Do you mean where you're using that R is a domain?

[crystal vale]

Yes

[rocky cloak]

You have a polynomial
x^n + c1 x^n-1 + ... + cn = 0

Since C is algebraically closed this factors as
(x - a1)(x - a2) ... = 0
for complex numbers ai.

Now you have a product of things that are zero, so in a domain one of the factors must be zero. I.e. x is equal to one of the ai

You already looked at the example C[x]/(x^2)

here x^2 = 0 factors as (x-0)(x-0), but x is not equal to 0.

[crystal vale]

I see

[crystal vale]

Yes

[chilly ocean]

there is another standard question, that evey finite dimensional vector space over a field which is also an integral domain is again a field

[crystal vale]

Yes I saw this question on MSE

[rocky cloak]

Or every artinian domain is a division ring

[chilly ocean]

how

[rocky cloak]

The proof is essentially the same.

Consider the chain of ideals generated by (x^n) to conclude that x is a unit

[chilly ocean]

That was done in class i forgot

[tardy hedge]

Greetings my friends. I will now probably be living in the combinatorial structures chat since I've now begun learning about my thesis topic

well. there is still algebra involved so i will prolly ask here too

i am doing combinatorial algebra stuff

[thorn jay]

Betraying us i see

Good luck

[tardy hedge]

thank u sir

true that man

I am learning about stanley-reisner theory atm

[thorn jay]

I always have trouble with combinatorics omfg

If its one branch of math I'll fail its that

[barren sierra]

What's the topic?

All the combo haters in this chat are missing out on some of the most beautiful math out there in algebraic combinatorics 😔

And what text are you using?

[tardy hedge]

My supervisor wants to look into expanding the ideas presented in the paper glicci simplical complexes by nagel and romer

[tardy hedge]

Rn i am looking through this review paper thats called a survey of stanley reisner theory and im also looking through monomial ideals by herzog and hibi

[barren sierra]

I've been meaning to look at that text, seems useful to keep on my shelf

Can you link the survey?

[tardy hedge]

Sure

[barren sierra]

I've seen Stanley Reisner theory come up in some stuff but idk what it is

[tardy hedge]

https://math.okstate.edu/people/mermin/papers/A_survey_of_Stanley-Reisner_theory.pdf

[barren sierra]

Beautiful

Thanks

[thorn jay]

I say the same thing about universal algebra but noooo

[barren sierra]

Yea but have you considered you're wrong and I'm right?

Most people tend to forget to consider this fact but it really is essential

[thorn jay]

I have, and rejected the thought as it was too silly

[elfin wraith]

Monomial ideals is really nice, but holy shit does it get hard fast haha

[ashen hill]

are there a source of sloutions

to the exercises

in artins algebra second ed

[glad osprey]

How do I prove this btw? I have a hunch that you could embed KL in the tensor product of K and L, which might have degree nk over F, but I know very little about tensor products, so I'm not sure

[rocky cloak]

What I said was slightly wrong actually.
There's a surjection from the tensor product to KL, so the degree is less than the product, but it doesn't necessarily divide it.

It will be a multiple of the lcm though

[glad osprey]

I see, thanks

[weak prawn]

Hey, need some help trying to classify groups.
I have the group of unities of $\mathbb Z_{21}$, which is of order 12 and is {1,2,4,5,8,10,11,13,16,17,19,20}. According to the fundamental theorem of finitely generated abelian groups I know this isomorphic to either $\mathbb Z_4 \cross Z_3$ or $\mathbb Z_2 \cross \mathbb Z_2 \cross \mathbb Z_3$, but how do I know which of them is isomorphic to the group above?

[cloud walrus]

Michael

[coral spindle]

Well OK

You can distinguish G = Z_4 x Z_3 and H = Z_2 x Z_2 x Z_3 by looking at the presence of an element of order 4. The group G has an element of order 4, the group H does not.

But this is easier still. Z_21 as a ring is isomorphic to the ring Z_3 x Z_7 by the Chinese remainder theorem. Now exercise: the group of units (R x S)* of the ring R x S is equal to (not merely isomorphic to!) R* x S*.

We understand well the group of units of Z_p where p is a prime. It is Z_{p-1}. Perhaps you know this -- it is not entirely straightforward to prove.

So in fact I can tell you that the group of units of Z_21 is isomorphic to Z_2 x Z_6 which in turn is isomorphic to Z_2 x Z_2 x Z_3.

No inspection of the group was necessary.

[weak prawn]

Huh, interesting. Didn't know that. And Z_6 is isomorphic to Z_2 x Z_3 since 2 and 3 are coprimes right?

[coral spindle]

That's right

That's in fact another application of the Chinese remainder theorem, just restricted to groups rather than rings

[weak prawn]

I see, nice

[weak prawn]

But if we want to use this way, do you often do it by just looking at the orders of different elements and seeing if they're structurally alike that way?

[coral spindle]

That is a quick and easy way to distinguish the groups but

Do I often do it that way? No haha

idk if computer algebra systems do

[weak prawn]

Hehe alright, do you have some other neat ways of distinguishing them?

[coral spindle]

This is pretty much the only way if you're just given them as a Cayley table

[weak prawn]

Okay well ty for answers!

[rocky cloak]

Basically the same idea, but the latter has more elements of order 2 than the the former.

It may or may not be easier to count solutions to x^2 = 1, than determine if there is an element of order 4

[coral spindle]

I guess it depends on which elements are listed first! :P

[next obsidian]

Why are you blue now

You joined them?

[vast stump]

them (derogatory)

[chilly ocean]

does there exist a 7*7 matrix such that the space of commuting matrices has dimension 6

[coral spindle]

When I woke up I was blue dabba dee dabba dai

[rocky cloak]

There does not.

Break the corresponding module up into cyclic module. Then the endomorphism ring of each summand is its dimension, so if you just look at the endomorphisms that preserve the decomposition that's already 7-dimensional.

Then you can have more if there are homomorphisms between the summands

[chilly ocean]

what do you mean by break module into cyclic module, is it direct sum

[rocky cloak]

Yes

[chilly ocean]

decomposition into cyclic module V= Rx_1 +Rx_2...+Rx_n end endomorphism of directic sum is direct sum of individual, using this we are getting dim 7

[rocky cloak]

The endomorphism ring of a direct sum decomposes as

Sum_i,j Hom(Rx_i, Rx_j)

But if we just count the i=j part that already gets us to 7

[chilly ocean]

oh yes

this result is nice

[proper jolt]

hello for this result does anyone know where or not

the statement would still work if we changed T = \phi_s^{-1}(T') instead of towards T'= \phi_s{T} as stated originally

im thinking that might not work because T'^{-1} (S^-1M) might not even be a T^-1 R module

is that correct?

[crude herald]

the isomorphism may fail because defining T from T' does not guarantee the necessary condition S ⊆ T (equivalently, φ_S(S) ⊆ T') required for the standard proof.

[proper jolt]

I'm still keeping the explicit condition that S \subset T

[median ether]

(from dummit,foote) i want to build a statement like this to prove it by induction: \
Prop. If f = (c1)(c2), $|f| = 2$ and $(c1) \cap (c2)$, then order of f = 1. \
f $\in S_\text{n}$ - product of disjoint cycles, seems not legitimate to use f = (c1)(c2). but if (c1)(c2) is not f, then how can we define an order?

[cloud walrus]

shdvv

[vocal pebble]

what do you mean by "and (c1) \cap (c2).."? that they intersect nontrivially?

[coral spindle]

Not sure about that proposition you've stated! Are you saying that e.g. (1 2 3)(4 5) is of order 1, i.e., is the identity?

Yeah any as pointed out you don't actually say what that should be equal to, though I apparently imagined it meant empty intersection

[vocal pebble]

either way your proposition is quite confusing because you say at the start |f| = 2 then at the end you say order f = 1

if |f| means the order of f, then that statement clearly cant be true

[chilly ocean]

if finitete intersection of maximal ideal is contained in some maximal is it true at least one max ideal od collection is equal to the ideal in which intersection is contained

[rocky cloak]

If a product of ideals is contained in a prime ideal, then one of the ideals is contained in the prime ideal.

Maximal ideals are prime and product is contained in intersection. So the answer should be yes

[coral spindle]

Realising I misread 🤦‍♂️

[chilly ocean]

can it be prooven using chinese remainder therem for maximal ideals

[rocky cloak]

Sure

[chilly ocean]

so image of maximal ideal is of the form R/M_1* R/M_2 .. 0... R/M_n

onto map so image of maximal ideal is maximal

[tardy hedge]

in a polynomial ring R[x1, x2, ... xn], for lets say R a domain, what are the prime ideals? I guess those generated by irreducible polynomials would still be right? And any ideal generated by some variables (xi : i in 1 to n)

Havent really thought about that much because I havent really worked with polynomial rings with more than one variable, and for the last while its just been polynomial rings over fields because I was doing galois theory

[south patrol]

I mean this is just super hard in general

[tardy hedge]

Yeah, I guess I just need to see more examples in general

[south patrol]

But checking specific examples is easy usually

Irreducible elements are not always prime though

[tardy hedge]

oh shoot

irreducibles are prime only in a UFD is it?

[south patrol]

It is fine here if R is a UFD (so R[x1,..,xn] is too)

[tardy hedge]

Ok

[south patrol]

Well I don't think it implies UFD

[tardy hedge]

In a domain where factorization into irreducibles is possible .... irreducible implies prime implies its a UFD?

i think i read that in my number theory notes

from my prof

[south patrol]

Yes, you need factorisation to be possible

And then this is true

Amusingly I went over the proof of this when presenting part of a number theory class recently lol

[tardy hedge]

oh nice haha

[rocky cloak]

I think you can describe it like, take P a prime ideal in R, then look at the algebraic closure of the field of fractions of R/P, let's call it K.

Then you get a map R -> K by sending the xi to arbitrary elements. I think they should correspond to the same ideal iff they have the same Galois orbit.

So then you're looking at the Galois orbits of K^n

This still misses some prime ideals though, since the xi don't need to map to something algebraic

... seems like it would be hard to give a good description though

[glad osprey]

Why does the proof that Gal(R/Q) is trivial not work for Gal(Q-bar/Q)? AFAICT the only ingredients is that any automorphism fixes Q and preserves order, plus the fact that Q is dense in R. This is true for Q-bar too, so what fails?

[delicate orchid]

what order precisely is being preserved here? Q-bar isn't a subset of R

[glad osprey]

omg

thanks

thought it was a subset of R for a weak second

[delicate orchid]

a model structure on seconds

[glad osprey]

but wait, what about Gal(Q(sqrt(2)/Q)? Why isn't that trivial by the same argument?

[delicate orchid]

because I can send sqrt(2) to -sqrt(2) so there is an automorphism

is Q dense in Q(sqrt(2))? what's the topology here

[glad osprey]

hmm... the subspace topology of R?

[delicate orchid]

well that's YUCK

[glad osprey]

surely that can't be the part of the argument that fails. Q(sqrt(2)) has a well defined order, right?

[delicate orchid]

I think this is the part where I just "Gal(R/Q) is trivial because Aut(R) is trivial"

[glad osprey]

isn't Galois groups and automorphism groups pretty much literally the same?

atleast over the prime field

[delicate orchid]

well yeah they're both trivial

oh I see what you mean. Yes this holds more generally

[glad osprey]

I still don't see which part of the argument fails I can't do any more math until I figure this out

[delicate orchid]

well it's pretty clear that for Q(sqrt(2)) the part that fails is that automorphisms have to be order preserving

so peep that part REALLY closely

[glad osprey]

ah, I see it: in R a number is positive iff it is a square, which is not true in Q(sqrt(2)). So an automorphism sends squares to squares, but not necessarily positives to positives

thanks

[tough raven]

Can you find an upper bound on this order, by finding a finite number of words in x, y such that every other word must be equal in G to one of them?

For example, how can you "simplify" xyxxyyxyxx?

[ivory ore]

well yeah i was thinking of words, i didnt see that i pressed the enter already

alr i would try a bit on this path

[ivory ore]

is there any specificity in the way you creawted that "xyxxy...." or it was just random word

[sturdy spear]

does it look fine?

[ivory ore]

Using the relation $(xy)^3 = 1$, we can derive rewriting rules such as $yxy = x^2 y x^2$ and $yx^2y = x y x$

[cloud walrus]

longboard kayak

[sturdy spear]

for the last reasoning if 0< z < n/d then n/d | z was impossible hence n/d < z for all z such that (x^k)^z=1

[ivory ore]

any word in $G$ may be written in the form $x^i$ or $x^i y x^j$ for $0 \le i,j < 3$. There are $3$ elements of the first type and $9$ of the second, yielding at most $12$ distinct elements.

[cloud walrus]

longboard kayak

[ivory ore]

chat got fragmented

this sounds?

[serene dune]

problem books?

[worthy solar]

How do I prove cos(2pi/5) is irrational so that I know Q(cos(2pi/5)) is not the same field as Q

The question was for finding the splitting field of x^5+1 and all it's subfields

[tough raven]

Mostly random, but I tried to make it not simplify too much.

[chilly ocean]

Cos(2 π/5 ) is x+ x^-1/2 where x is primitive 5th root of unity so is x^-1

[worthy solar]

But why does that matter. I was thinking about the complex conjugates but wasn't sure what to do with it

[chilly ocean]

If Q(x+x^-1)= Q then x+x^-1 is some rational

[worthy solar]

Agreed

[chilly ocean]

So x^2-qx+1 =0 minimal polynomial for x

[chilly ocean]

But the degree of minimal polynomial for x is phi(5) that is 4

[chilly ocean]

This result you can find in cycloctomic extensions in dummit foote

[worthy solar]

Where x = e^(i2pi/5)?

[worthy solar]

That's what I was afraid of

Didn't really discuss cycloctomic polynomials

[chilly ocean]

Yes

[worthy solar]

Hence was trying to avoid these thms 😭

[chilly ocean]

That's a part of field theory , you can use that

[worthy solar]

The thing is it kinda trivializes some of the x^p-1 questions, no?

I mean at least knowing what my galois group should be

[chilly ocean]

And this type of extensions usually seen in cycloctomic extensions part

[worthy solar]

Oh I guess that theorem was actually for all n

I suppose I'll use it and move on

[chilly ocean]

Which theorem

[worthy solar]

What the Galois group corresponds to the splitting field x^n-1

,w primitive root

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by brandon7716

[worthy solar]

Is this the same as a cyclic generator?

Of U(p)

Or I guess people say (Z/pZ)^x

[chilly ocean]

Yes

[chilly ocean]

Oh man is this a new way to say cos(2π/7) is irrational

[worthy solar]

Which there is some know result about which n for has a cyclic group for U(n)

[worthy solar]

My book just uses that notation for invertible residues mod n

[chilly ocean]

When n is 2,4 ,p^n,2p^n

Where p is odd prime

[worthy solar]

Yes

[worthy solar]

I meant to say cos(2pi/5) but if I can just use the properties of cycloctomic polynomials then I basically skip all of that

[chilly ocean]

Yes

[rocky cloak]

I guess one can probably to some trigonometry to deduce it's equal to (sqrt(5) - 1)/4

If you want to avoid Galois theory

[worthy solar]

Well I need to use Galois theory because I have to construct all the subfields. However I was am starting to realize idk how to show some polynomials are minimal and or not. Also a struggle with knowing which permutations of the roots are valid

The only Galois theory I have is basically finding the subfields of a field extension is related to the Galois group where we analyze the group automorphisms of the field extension

[rocky cloak]

Can you use Eisenstein?

[worthy solar]

Yes but my understanding there is a difference between minimal and irreducible

Like if one tacks on another complex root to your polynomial

[rocky cloak]

Over a field a polynomial is minimal iff it is irreducible

Anyway, you can notice that x^5 + 1 and (-x-1)^5 + 1 has the same splitting field.

The latter is
-x^5 - 5x^4 - 10x^3 - 10x^2 - 5x
which after you divide by x is irreducible by Eisenstein.

[worthy solar]

But can't I just have say like (x-r1)(x-r2)(x-r1) and add say another complex root? Or is the issue that it might be reduce from an overall degree 4 to 2 degree 2s

[rocky cloak]

I don't understand what you're suggesting

[worthy solar]

Or maybe that's not possible either for Q[x] if the complex roots come in pairs

[worthy solar]

Basically like Q(i) but then instead of suggesting (x^2+1) I say (x^2+1)(x^2-2)

The latter reduces into 2 degree 2 polynomials

[rocky cloak]

And are you suggesting the second one minimal or irreducible? (it's neither)

[tough raven]

Isn't X^5 + 1 = Phi_2 Phi_10, so that the spltting field is that of Phi_10, i.e., ℚ(zeta_10)?

[rocky cloak]

Yup

[worthy solar]

Well I just showed it is reducible I guess and so neither because we know it should be degree 2

[tough raven]

Both of those are polynomials satisfied by i (and there are more), but x^2 + 1 is the only (up to scaling by a constant) irreducible one and is also the minimal polynomial of i.

[worthy solar]

Yeah.

Anyways I was basically thinking of how the degree 5 polynomial could reduce and gets it's minimal polynomial for Q(w). I mean we know (x-1) is linear factor but 1 is already in Q so we are left with the degree 4 polynomial

But that requires me to show x^4+x^3+x^2+x+1 does not reduce

Over Q